来自北大算法课的Leetcode题解：46. 全排列

代码仓库：Github | Leetcode solutions @doubleZ0108 from Peking University.

• 解法1(T83% S98%)：直接调用python库函数 itertools.permutations(nums, len(nums))
• 解法2(JS T17% S5%)：回溯，参数分别为parts数组和visited集合，当parts的长度等于原数组长度则添加到结果中，visited用于存储哪些数字已经用过了，从原数组中排除这些用过的，就是当前位可以添加并继续递归的

var permute = function(nums) {
    var res = [];
    var sets = new Set(nums)
    var deepin = function(parts, visited) {
        if (parts.length == nums.length) {
            res.push(parts);
            return;
        }
        
        for (var item of [...nums].filter(x => !new Set(visited).has(x))) {
            deepin(parts.concat(item), visited.concat(item));
        }
    };
    deepin([], []);
    return res;
};

class Solution:
    def permute(self, nums: List[int]) -> List[List[int]]:
        return list(itertools.permutations(nums, len(nums)))